Integrals are equal? How? Why are these integrals the same?
$$ \int_0^1\bigg(\frac{-1}{\ln^3(x)}\bigg)\exp\bigg(\frac{1}{\ln(x)}\bigg)~dx=\int_0^1 \bigg(-\ln(x)\bigg)\exp\bigg(\frac{1}{\ln(x)}\bigg)~dx=2K_2(2) $$
Where the $K_2$ is a modified Bessel function.
Is there some property that describes when this happens?
I think it has to do with the functions being in the same general form, but I can't pinpoint the specifics.
Thanks.
 A: We can show that the difference between them is $0$ which shows how they are equal.
$$\int_0^1 \exp\left(\frac{1}{\ln x}\right)\left(\ln x-\frac{1}{\ln^3 x}\right)dx\overset{\ln x \to x}=\int_{-\infty}^0 e^{x+1/x}\left(x-\frac{1}{x^3}\right)dx$$
$$=\int_{-\infty}^0 e^{x+1/x}\left(1-\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)dx=\int_{-\infty}^0 \left(e^{x+1/x}\right)'\left(x+\frac{1}{x}\right)dx$$
$$\overset{IBP}=-\int_{-\infty}^0 e^{x+1/x}\left(1-\frac{1}{x^2}\right)dx=-e^{x+1/x}\bigg|_{-\infty}^0=0$$

This can be further generalized. A more direct approach than the one from above is as follows:
$$I(n)=\int_0^1 \exp\left(\frac{1}{\ln x}\right)\left(\ln^{n-1} x-\frac{1}{\ln^{n+1}x}\right)dx\overset{\ln x\to x}=\int_{-\infty}^0 e^{x+1/x}\left(x^{n-1}-\frac{1}{x^{n+1}}\right)dx$$
$$\overset{x\to 1/x}=\int_{-\infty}^0 e^{x+1/x}\left(\frac{1}{x^{n+1}}-x^{n-1}\right)dx\Rightarrow 2I(n)=0$$
$$\Rightarrow \int_0^1 \exp\left(\frac{1}{\ln x}\right)\ln^{n-1} xdx=\int_0^1 \exp\left(\frac{1}{\ln x}\right)\frac{1}{\ln^{n+1}x} dx$$
